Question

Find sum of all three digit numbers which are dividible by three but not divisible by 5

Answer 1

An arithmetic progression is a sequence of number of the form a,a+d,a+2d,a+3d,..,a+nda,a+d,a+2d,a+3d,..,a+nd, that is every two consecutive numbers differ by the same value called dd.

The formula for the sum of the first nn terms of a progression starting at aa with common difference dd is n2[2a+(n−1)d]n2[2a+(n−1)d].

To solve your problem we need to add all the numbers from 100 to 999 and then subtract the numbers that can be divided exactly by 33.

The sum of the numbers is the sum of the first 900900 terms of a progression with a=100a=100 and d=1d=1, that is 9002(200+899)=4945509002(200+899)=494550

Now we need to calculate the sum of the first ⌈999−1003⌉=300⌈999−1003⌉=300 terms of the sequence with a=102a=102 and d=2d=2, that is 3002(2(102)+3(299))=1651503002(2(102)+3(299))=165150, the result is then 494550−165150=329400494550−165150=329400

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